Which one of the following is not a basic solution of system of linear equation

x+ 2x+ x= 4

2x+ x+ 5x= 5

  1. x= -1, x= 2, x3 = 1
  2. x= 2, x= 1
  3. x= 5, x= -1
  4. x= 5/3, x= 2/3

Answer (Detailed Solution Below)

Option 1 : x= -1, x= 2, x3 = 1

Detailed Solution

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The correct answer is : option 1

Key Points

Concept of Basic Solution:

  • A basic solution to a system of linear equations in matrix form (Ax = b) is obtained by setting (n - m) variables to zero, where:
    • n = total number of variables
    • m = number of equations
  • Here, we have 2 equations (m = 2) and 3 variables (x₁, x₂, x₃), so:
    • We must set (3 - 2) = 1 variable to zero to get a basic solution.
  • If all variables are non-zero, it is not a basic solution.

Now, let’s analyze each option:

 Option 1) x₁ = -1, x₂ = 2, x₃ = 1

  • All variables are non-zero → Not a valid basic solution.
  • Check equation 1: -1 + 2×2 + 1 = -1 + 4 + 1 = 4 ✔
  • Check equation 2: 2×(-1) + 2 + 5×1 = -2 + 2 + 5 = 5 ✔
  • This is NOT a basic solution because no variable is zero.

 Option 2) x₁ = 2, x₂ = 1, x₃ = 0

  • x₃ = 0 (one variable is zero) → could be a basic solution.
  • Equation 1: 2 + 2×1 + 0 = 4 ✔
  • Equation 2: 2×2 + 1 + 5×0 = 4 + 1 = 5 ✔
  • Valid basic solution.

 Option 3) x₁ = 5, x₂ = 0, x₃ = -1

  • x₂ = 0 → One variable zero → Candidate for basic solution.
  • Equation 1: 5 + 0 + (-1) = 4 ✔
  • Equation 2: 2×5 + 0 + 5×(-1) = 10 - 5 = 5 ✔
  • Valid basic solution.

 Option 4) x₂ = 5/3, x₃ = 2/3

  • Need to find x₁:
  • Equation 1: x₁ + 2×(5/3) + (2/3) = 4 → x₁ + 10/3 + 2/3 = 4 → x₁ = 4 - 12/3 = 0
  • x₁ = 0 → One variable is zero → Possible basic solution
  • Equation 2: 2×0 + 5/3 + 5×(2/3) = 5/3 + 10/3 = 15/3 = 5 ✔
  • Valid basic solution.

✅ Final Answer:

Option 1) x₁ = -1, x₂ = 2, x₃ = 1 is NOT a basic solution because all variables are non-zero.

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